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These questions are really mean but if given in the exam give ALOT of marks and whilst scary at first follow a very general pattern. First I would take either equation and re-arrange for the lone variable...

The general equation of a circle is (x-a)^2 + (y-b)^2 = r^2 where the centre is (a,b) and radius r.
Therefore the equation of this circle is (x-3)^2 +(y-4)^2 = 4^2 ...or 16
Sometimes its easie...

Use substitutionlet u = cos(x)du = - sin(x)dxint(tan(x)) dx = int(sinx/cosx) dx = - int(1/u) du = - ln(u) + c= ln(secx) + c

We will use the cross product of these two vectors to help us find the perpendicular vector. The cross product works by finding a vector which has no effect on the position of a point with respect the the...

2 Parts to this question - requires a complimentary function (treating equation as if right-hand side (RHS) = 0) and a particular integral (the general solution for where the RHS = 26sin(3x) )
1) The...